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erik [133]
3 years ago
13

Quantization of energy lab Hypothesis: Make a prediction that describes the relationship between the composition of an unknown s

ubstance and its emission spectrum. For example, “If each element can be identified by its ______________, then the ____________ of an unknown star can be determined." variables: independent - dependent- control-
Chemistry
2 answers:
elixir [45]3 years ago
8 0

Answer:

If each element can be identified by its spectrum then the composition of an unknown star can be determined

Explanation:

The chemical nature of the elements is that they absorb specific wavelength of light depending on their atom. By spectral analysis of the spectrum of emitted light by a body, the body's composition can therefore be determined. As such in order to determine the composition of distant bodies such as planets, stars and other celestial bodies scientists usually make use of spectroscopy.

bezimeni [28]3 years ago
7 0

Answer:

If each element can be identified by its spectrum then the composition of an unknown star can be determined

Explanation:

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PLEASE HELP DUE IN EXACTLY 15 mins!! i will give you branliest
Deffense [45]

2032533 \sq22222rt[2]{?}

4 0
3 years ago
If an ideal gas has a pressure of 5.49 atm, a temperature of 88.78 °c, and has a volume of 22.03 l, how many moles of gas are in
kirza4 [7]
N = (PV)/RT
(T = 88.78 + 273 = 361.78K)
(R = 22.4/273 = 0.082)
= (5.49 x 22.03)/(0.082 x 361.78) = ?
Put it into the calculator. It's hard to do that on a mobile phone.
7 0
3 years ago
A 1.59 mol sample of Kr has a volume of 641 mL. How many moles of Kr are in a 4.41 L sample at the same temperature and pressure
Marina86 [1]

Answer:

The correct answer is 10.939 mol ≅ 10.94 mol

Explanation:

According to Avogadro's gases law, the number of moles of an ideal gas (n) at constant pressure and temperature, is directly proportional to the volume (V).

For the initial gas (1), we have:

n₁= 1.59 mol

V₁= 641 mL= 0.641 L

For the final gas (2), we have:

V₂: 4.41 L

The relation between 1 and 2 is given by:

n₁/V₁ = n₂/V₂

We calculate n₂ as follows:

n₂= (n₁/V₁) x V₂ = (1.59 mol/0.641 L) x 4.41 L = 10.939 mol ≅ 10.94 mol

5 0
2 years ago
Classify these definitions as that of an Arrhenius acid, an Arrhenius base, or other. Arrhenius acid definition Arrhenius base d
stealth61 [152]

Answer:

Explanation:

A substance that produces an excess of hydroxide ion (-OH) in aqueous solution.

       This is an arrhenius Base

According to the arrhenius theory, a base is a substance that combines with water to produce excess hydroxide ions, OH⁻ in an aqeous solution. Examples are :

  • Sodium hydroxide NaOH
  • Potassium hydroxide KOH

A substance that produces an excess of hydrogen ion (H+) in aqueous solution

      This is an arrhenius Acid

An arrhenius acid is a substance that reacts with water to produce excess hydrogen ions in aqueous solutions.

Examples are;

  • Hydrochloric acid HCl
  • Hydroiodic acid HI
  • Hydrobromic acid HBr
5 0
3 years ago
You will need to prepare 12 mL of 25% Sodium Phosphate Buffer (pH 4) solution for Activity 2. What volume of the stock Sodium Ph
zhuklara [117]

Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>

The process of preparing solutions from stock solutions of higher concentration is known as dilution.

Dilution is done with the aid of the dilution formula given below:

  • C1V1 = C2V2

where

  • C1 is the concentration of stock solution
  • V1 is the volume of stock solution required to prepare a diluted solution
  • C2 is the concentration of the diluted solution prepared
  • V2 is the final volume of the diluted solution

From the data provided:

C1 is not given

V1 is unknown

C2 = 25%

V2 = 12 mL

  • Assuming C1 is 50% solution

Volume of stock, V1, required is calculated as follows:

V1 = C2V2/C1

V1 = 25 × 12 /50

V1 = 6 mL

Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

Learn more about dilution formula at: brainly.com/question/7208546

6 0
2 years ago
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